2. When iron (II) hydroxide is mixed with phosphoric acid, iron (II) phosphate p

Question

2. When iron (II) hydroxide is mixed with phosphoric acid, iron (II) phosphate precipitate results as follows: Fe(OH): (aq) + HPO4 (aq) Fe( PO(s) + 110 (1) a. Balance the reaction equation b. If 3.20 gof Fe(OH) is treated with 2.50 g of phosphoric acid, what is the limiting reagent and what is the reactant in excess? What mass of Fe (PO) will be produced? d. c. Calculate the percent yield if 3.99 g of Fe(POis collected 3. Consider the following reaction of benzene and bromine to produce bromobenzene and hydrobromic acid: C6H6 + Br2 C6H,Br + H Br If 42.1 g of C&Hs; reacts with 73.0 g of Br, what is the limiting reagent and what is the reactant in excess? What is the theoretical yield of C&HsBr;? The graduated cylinder you collect your product in weighs 11.7g, if the total mass of C&HsBr; and your graduated cylinder is 75.3g, what is the percent yield of your collected product? a. b. c.

Explanation / Answer

2

a.

Three molecules of Fe(OH)2 reacts with two molecules of H3PO4 to precipitate one molecule of Fe3(PO4)2 and six molecules of H2O. Thus, balanced equation can be written as follows:

3Fe(OH)2(aq) + 2H3PO4(aq)   ------->   Fe3(PO4)2(s) + 6H2O(l)

b.

Molar mass of Fe(OH)2 is 89.86 g/mol. Calculate number of moles of Fe(OH)2 in 3.20 g.

Number of moles of Fe(OH)2 = weight / molar mass

Number of moles of Fe(OH)2 = (3.20 g) / (89.86 g/mol)

Number of moles of Fe(OH)2 = 0.03561 mol

Therefore, number of moles of Fe(OH)2 in 3.20g are 0.03561 mol.

Molar mass of H3PO4 is 97.994 g/mol. Calculate number of moles of H3PO4 in 2.50 g.

Number of moles of H3PO4 = weight / molar mass

Number of moles of H3PO4 = (2.50 g) / (97.994 g/mol)

Number of moles of H3PO4 = 0.02551 mol

Therefore, number of moles of H3PO4 in 2.50 g are 0.02551 mol.

According to balanced reaction equation, 3 moles of Fe(OH)2 reacts with 2 moles of H3PO4. Thus, calculate number of moles of H3PO4 that can react with 0.03561 moles of Fe(OH)2.

Number of moles of H3PO4 required = (2/3) x 0.03561 moles of Fe(OH)2

Number of moles of H3PO4 required = 0.02374 mol

Therefore, given 0.03561 moles of Fe(OH)2 can react with maximum 0.02374 moles of H3PO4. However, given number of moles of H3PO4 are 0.02551moles, which is in excess.

Hence, limiting reagent is Fe(OH)3 and excess reactant is H3PO4.

c.

According to balanced equation, three moles of Fe(OH)2 gives one mole of Fe3(PO4)2 and limiting reagent is Fe(OH)2. Thus, calculate number of Fe3(PO4)2 that can be obtained theoretically from given 0.03561 moles of Fe(OH)2.

Number of moles of Fe3(PO4)2 = (1/3) x number of moles of Fe(OH)2

Number of moles of Fe3(PO4)2 = (1/3) x 0.03561 moles

Number of moles of Fe3(PO4)2 = 0.01187

Thus, theoretically 0.01187 moles of Fe3(PO4)2 are obtained.

Molar mass of Fe3(PO4)2 is 357.48 g/mol. Thus, calculate mass of 0.01187 moles of Fe3(PO4)2.

Mass of 0.01187 moles of Fe3(PO4)2 = (0.01187 mol) x (357.48 g/mol)

Mass of 0.01187 moles of Fe3(PO4)2 = 4.243 g

Therefore, theoretically mass of Fe3(PO4)2 produced in given reaction will be 4.243 g.

d.

Theoretical yield is 4.243 g and practical yield is 3.99 g. Thus, calculate percent yield as follows:

Percent yield = (practical yield / theoretical yield) x 100

Percent yield = (3.99 g / 4.243 g) x 100

Percent yield = 94.0 %

Therefore, percent yield is 94.0 %

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3

a.

The balanced equation of reaction of benzene with bromine is given below:

C6H6   + Br2 -------->    C6H5Br   +   HBr

Thus, one mole of benzene(C6H6) reacts with one mole of bromine(Br2) to give one mole of bromobenzene(C6H5Br) and one mole of HBr.

Molar mass of benzene is 78.11 g/mol. Calculate number of moles of benzene in 42.1 g.

Number of moles of benzene = weight / molar mass

Number of moles of benzene = (42.1 g) / (78.11 g/mol)

Number of moles of benzene = 0.539 mol

Therefore, number of moles of benzene in 42.1 g is 0.539 mol.

Molar mass of bromine is 159.8 g/mol. Calculate number of moles of bromine in 73.0 g.

Number of moles of bromine = weight / molar mass

Number of moles of bromine = (73.0 g) / (159.8 g/mol)

Number of moles of bromine = 0.457 mol

Therefore, number of moles of bromine in 73.0 g is 0.457 mol.

Thus, 0.457 moles of bromine can react with 0.457 moles of benzene. However, moles of benzene are 0.539 mol, which is in excess. Therefore, bromine is the limiting reagent and excess reactant is benzene.

b.

Limiting reagent is bromine. Since, one mole of bromine gives one mole of bromobenzene. Therefore, 0.457 moles of bromine will give 0.457 moles of bromobenzene.

Molar mass of C6H5Br is 157 g/mol. Thus, calculate weight of 0.457 moles of bromobenzene.

Weight of bromobenzene = moles x molar mass

Weight of bromobenzene = (0.457 mol) x (157 g/mol)

Weight of bromobenzene = 71.7 g

Therefore, theoretical yield of bromobenzene is 71.7 g.

c.

Actual weight of bromobenzene = Total weight - wt. of the cylinder

Actual weight of bromobenzene = 75.3 g - 11.7 g

Actual weight of bromobenzene = 63.6 g

Therefore, actual weight of bromobenzene obtained is 63.6 g. Theoretical yield is 71.7 g. Calculate percent yield.

Percent yield = (practical yield / theoretical yield) x 100

Percent yield = (63.6 g / 71.7 g) x 100

Percent yield = 88.7 %

Therefore, percent yield of collected product is 88.7 %

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